i thinkk it it B
search on net, there ar elots of info about inequalities

My answer is D.

St1 is sufficient. The expression evaluates to true for both (x, y) = (2, 1) and (2, -3).
St2 is sufficient. xy < 0, therefore (x > 0 AND y < 0) OR (x < 0 AND y > 0). Trying (2, -3) and (-2, 3), the expression evaluates to true.

EDIT: Since you asked for tips, I think the best way is to just pick numbers intelligently which satisfy the statements when evaluating them.

EDIT: The correct answer is B. When evaluating St1, I failed to account for y = 0 for which the inequality evaluates to false. I need to take my own advice. Make sure you don't assume anything that is not explicitly given.

You can use numbers to test such cases. Just test two cases where you get a No and a Yes and you are good to eliminate that statement

In this case. For statement 1, consider
x=2, y=1, substituting these values into the equation, we get
Is 1>1, No
x=2 y=-1
Is 3>-1, Yes. We can eliminate answers A and D

For statement 2, either x>0 and y<0 or x<0 and y>0. Let's try some cases
x=2, y=-2
Is 4>0, Yes
x=-2,y=1, Yes. You will see that the equation on the left will always be the absolute sum of x and y.
So answer is B